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Question: A rectangular piece of cardboard is \[40\] inches wide and \[50\] inches long. Square \[5\] inches o...

A rectangular piece of cardboard is 4040 inches wide and 5050 inches long. Square 55 inches on a side are cut out of each corner, and the remaining flaps are bent up to form an open box. The number of cubic inches in the box is
F. 12001200
G. 78757875
H. 60006000
I. 80008000
J. 1000010000

Explanation

Solution

While solving the question one must remember that the cut of 55 in. is done on both the sides of the length of the box as a rectangle has two lengths similarly, the cut is also done in the width as well leaving a total cut size of 5+5=105+5=10 in. and apart from the sides a bottom bend is also made for the box to be closed at the bottom giving a height of 55 in. making the volume of the box as:
Volume=Length×Breadth×Height=\text{Length}\times \text{Breadth}\times \text{Height}

Complete step by step solution:
According to the question given, the length, width and the height of the box is given as 4040 in. wide and 5050 in long, now the box is being folded up by 55 in. on all sides changing the rectangle base dimension of the box by 55 in. on all sides giving the base of the box new dimensions of:
The width of the box is 4040 in. and the sides folded are on both sides making the width of the base of the box as:
⇒(40−(5+5))=30\Rightarrow \left( 40-\left( 5+5 \right) \right)=30 in.
The length of the box is 5050 in. and the sides folded are on both sides making the length of the base of the box as:
⇒(50−(5+5))=40\Rightarrow \left( 50-\left( 5+5 \right) \right)=40 in.

Now according to the diagram of the box, we have the dimensions of the box as width 3030 in., length4040 in. and height as 55 in. Now using these dimensions, we find the volume of the box as:
Volume=Length×Breadth×Height=\text{Length}\times \text{Breadth}\times \text{Height}
Volume=40×30×5=\text{40}\times \text{30}\times \text{5}
Volume=6000=6000 cubic in.
Therefore, the volume of the open box is 60006000 cubic in.

Note: The dimensions are subtracted twice because the box is folded 2 times in the length section and 2 times in the width section but the height will remain 55 in. only thereby subtracting twice from both the length and the width.